Seja:

\(f(x,y)=cos(x³+y²)\)

sua derivada parcial em relação a x é:

\(\frac{\partial }{\partial \:x}\left(cos\left(x³+y²\right)\right)\)

Vamos utilizar a regra da cadeia \(\frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\) com \(f=\cos \left(u\right),\:\:u=x^3+y^2\)

\(\frac{\partial \:}{\partial \:u}\left(\cos \left(u\right)\right)\frac{\partial \:}{\partial \:x}\left(x^3+y^2\right)\)

Mas:

\(\frac{\partial \:}{\partial \:u}\left(\cos \left(u\right)\right)=-\ sen \left(u\right)=-\ sen \left(x^3+y^2\right)\\ e\\ \frac{\partial \:}{\partial \:x}\left(x^3+y^2\right)=3x^2\)

Assim:

\(\frac{\partial \:}{\partial \:u}\left(\cos \left(u\right)\right)\frac{\partial \:}{\partial \:x}\left(x^3+y^2\right)=\left(-\ sen \left(x^3+y^2\right)\right)\cdot \:3x^2\)

Portanto:

\(\boxed{\frac{\partial }{\partial \:x}\left(cos\left(x³+y²\right)\right)=-3x^2\ sen \left(x^3+y^2\right)}\)